Year 12 Mathematics EAS 2 - Nulake 2.6 Sample.pdf · Year 12 Mathematics Contents ... solving...

Year 12Mathematics

Contents

uLake Ltdu a e tduLake LtdInnovative Publisher of Mathematics Texts

Robert Lakeland & Carl NugentAlgebraic Methods

x2 + 5x + 6 = 3

5 log 3

h = Vπr

2–

?• AchievementStandard .................................................. 2

• Expanding............................................................. 3

• Factorising............................................................. 6

• Exponents .............................................................. 12

• ChangingtheSubject ................................................... 16

• Logarithms............................................................. 21

• RationalExpressions.................................................... 29

• LinearEquations....................................................... 35

• LinearInequations...................................................... 41

• QuadraticEquations..................................................... 44

• QuadraticFormula...................................................... 49

• TheNatureoftheRootsofaQuadratic..................................... 53

• ExcellenceQuestionsforAlgebra.......................................... 56

• PracticeExternalAssessment............................................. 60

• Answers............................................................... 72

• OrderForm............................................................ 79

EAS 2.6

EAS 2.6 – Year 12 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent

2 EAS 2.6 – Algebraic Methods

Thisachievementstandardinvolvesapplyingalgebraicmethodsinsolvingproblems.

◆ ThisachievementstandardisderivedfromLevel7ofTheNewZealandCurriculumandisrelatedto theachievementobjectives ❖ manipulaterational,exponential,andlogarithmicalgebraicexpressions ❖ formanduselinearandquadraticequations.

◆ Applyalgebraicmethodsinsolvingproblemsinvolves: ❖ selectingandusingmethods ❖ demonstratingknowledgeofalgebraicconceptsandterms ❖ communicatingusingappropriaterepresentations.

◆ Relationalthinkinginvolvesoneormoreof: ❖ selectingandcarryingoutalogicalsequenceofsteps

❖ connectingdifferentconceptsorrepresentations ❖ demonstratingunderstandingofconcepts ❖ formingandusingamodel; andalsorelatingfindingstoacontext,orcommunicatingthinkingusingappropriatemathematical statements.

◆ Extendedabstractthinkinginvolvesoneormoreof: ❖ devisingastrategytoinvestigateorsolveaproblem

❖ identifyingrelevantconceptsincontext ❖ developingachainoflogicalreasoning,orproof ❖ formingageneralisation; andalsousingcorrectmathematicalstatements,orcommunicatingmathematicalinsight.

◆ Problemsaresituationsthatprovideopportunitiestoapplyknowledgeorunderstandingof mathematicalconceptsandmethods.Situationswillbesetinreal-lifeormathematicalcontexts.

◆ Methodsincludeaselectionfromthoserelatedto: ❖ manipulatingalgebraicexpressions,includingrationalexpressions ❖ manipulatingexpressionswithexponents,includingfractionalandnegativeexponents ❖ determiningthenatureoftherootsofaquadraticequation ❖ solvingexponentialequations(whichmayincludemanipulatinglogarithms) ❖ formingandsolvinglinearandquadraticequations.

Achievement Achievement with Merit Achievement with Excellence• Applyalgebraicmethodsin

solvingproblems.• Applyalgebraicmethods,

usingrelationalthinking,insolvingproblems.

• Applyalgebraicmethods,usingextendedabstractthinking,insolvingproblems.

Algebraic Methods 2.6



Factorising

There are four methods of factorising an expression. The flow chart below may be helpful in identifying the correct approach for a particular expression.

No Yes

ax2 + bx + c

No Yes

Does each termhave a common

factor?

Follow thecommon factor

approach.

Are there fourterms (2 pairs)which could be

grouped?

No Yes

Are there amaximum of

three terms whichform a quadratic?

Either yourexpression cannotbe factorised or

you have made awrong decision.

Is the coefficientin front of the x

term 1?2

No YesFollow the

complexquadratic (a 1)

approach.

Follow the simplequadraticapproach.

Follow thegrouping of pairs

of termsapproach.

≠

FactorisingRewritinganexpressionasaproductofitsfactorsiscalledfactorising.Theexpression4x+36couldbewrittenas4(x+9)becausewhenwemultiplyitoutweget4x+36.Factorisingisthereverseproceduretoexpanding.

Weonlyneedtofactoriseanexpressionifitconsistsofasum(ordifference)oftwoormoreterms.Wemustthenfactorisealloftheexpressionnotjustpartofit.

Weadoptdifferentfactorisingtechniquesfordifferenttypesofexpressions.

Common FactorsWhenwestudytheexpressionbelowweseethatallthetermsthatmakeituphaveafactorincommon,sotheappropriateapproachistoextractthehighestcommonfactor.Forexample,eachtermintheexpression

4x3y2+6xy5–20x2y7z

has2,x,andy2ascommonfactors.Weidentifythesebyfirststudyingthenumbersatthefront(coefficients)andseeingthat2isthehighestcommonfactor,thenwestudythexcomponent(x3,xandx2)andnotexisthehighestcommonfactorandfinallytheycomponent(y2,y5andy7)andnotey2isthehighestcommonfactor.

Wewritetheseontheoutsideofthebracketsanddividethemintothecomponentofeachterm.

4x3y2+6xy5–20x2y7z =2xy2(2x2+3y3–10xy5z)

2dividesinto4,6and20togive2,3and10.

xdividesintox3,xandx2togivex2,1andx.

y2dividesintoy2,y5andy7togive1,y3andy5.

Ourexpressionbecomes

4x3y2+6xy5–20x2y7z=2xy2(2x2.1+3.1.y3–10xy5z)

=2xy2(2x2+3y3–10xy5z)

(x + 1)(x – 4) = x2 – 3x – 4

Factorising

Expanding ©uLake LtduLake Ltd Innovative Publisher of Mathematics Texts

21EAS 2.6 – Algebraic Methods


Logarithms

Logarithms

Heusedtheprinciplethatinmultiplyingthesamebasenumbertoapowerorexponentweaddthepower. 8x32=23x25 =23+5 =28 =256Aproblemsuchas23.45x456.2isdifficultandtediousbyhandbutifthenumbersareabletobeturnedintopowersof10theproblemchangestoaddition.23.45x456.2=101.370x102.659 =104.029 =10690(4sf)Tablesoflogarithms(andthereverseantilogarithms)wereproducedsopeoplecouldjustlookuptheappropriatelogandaddthem(orsubtractfordivision).Wecontinuetolearnanduselogarithmsastheymakemanyproblemseasier.

A short history of LogarithmsLogarithmswereinventedbytheScottishmathematicianJohnNapier.

Hedidsobecausetheyeliminatedtheneedtomultiply(ordivide),asatthetimeallproblemshadtobedonebyhand.

Logarithmsareusedwhenwewishtofindthepoweraparticularbase(oftenbutnotalways10)israisedbytoequalthenumberweareconcernedwith.

Ingeneralforbaseb

x=by

thenlogbx=y

Inbase10thelogarithmofthenumber100isthepower10wouldberaisedbytoequal100.

100=102

log10100=2Onyourcalculatorthebuttongivesthebasetenlog.Ifwefindlog100usingthecalculatorwegettheanswer2.

Similarlyifwecalculatelog3usingthecalculatorwegettheanswer0.4771as

3=100.4771

log103=0.4771

Tocheckthisoutwefind100.4771onthecalculatorandto4significantfiguresitreturnstheanswer3.

Logswereintroducedtoremovetheneedformultiplicationanddivisionpriortotheinventionofcalculators(seethenotesontheright).Wecontinuetouselogarithmsastheymakethesolutionofsomeproblemsaloteasier.

Wemakeuseoftheequivalentstatementoflogsinsolvingmanyproblems.

y=bxisequivalentto

logby=x

Forexample,tofindwhatbasetwonumberhasalogof5wesubstituteintotheexpression

log2Z=5

impliesZ=25

Z=32

Ifthebaseisnotstatedthenbyconventionthebaseistakenasbase10.

logN=log10N

log

The base of a log must be positive and not equal to 1.For logb y = x

then y = bx

only if b > 0 and b ≠ 1.

Justification for b > 1. If we have a base b < 0 and we define c = |b| then b = (–1) x c. logby = x y = bx

y = ((–1) x c)x

y = (–1)x.cx

For whole number values of x the sign oscillates depending upon whether x is odd or even and when x is a fraction it implies we are taking a root of a negative number therefore we cannot have a log base that is negative.

Justification for b ≠ 1. If b = 1 then it impliesFor log1 y = xthen y = 1x

As 1x = 1 this equation has only a trivial solution of y = 1 for all x.

©uLake LtduLake Ltd Innovative Publisher of Mathematics Texts



Achievement–Usetheequivalentlogstatement(y=bxisequivalenttologby=x)tosolvethe following.

173. log525=A 174.log8512=B

175. log264=C 176.log20.5=D

177. log3G=3 178.log10H=3

179. logM32=5 180. logN10=1

181. logP1=0 182.logQ243=5

Merit–Solvethefollowing.

183. log2(4y+4)=5 184. log3(2x–1)=2

185. log51x+ 23⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟= 2 186. log2(x2)=6

187. logx4(x–1)=2 188. logz(7z–10)=2




Example

Solvetheequation

2x2–x=36

2x2–x=36 Rearrangesothattheequationequalszerofirst 2x2–x–36=0 Factorising (2x–9)(x+4) =0 Settingeachfactorequalto0gives (2x–9) =0 x =4.5 and (x+4) =0 x =–4 x =–4,4.5

Quadratic Equations

Quadratic EquationsIfweareabletofactoriseaquadraticequationthenfindingthesolutionsisstraightforward.

Ifwehaveaproductoftwofactorsthatequal0thenone(orboth)ofthefactorsmustequal0. A.B =0

theneither A =0orB=0

Thereforeforaquadraticequation

(x+3)(x+2) =0

theneither (x+2) =0

x =–2

or (x+3) =0

x =–3

Thesolutioniswrittenas x =–3,–2Ifthequadraticisnotalreadyfactorised,thenwemustfactoriseitforthismethodtowork.

Be on the look out for quadratic equations disguised as something else.

A quadratic equation is a polynomial where the highest power of the unknown is two. A example is 3x2 + x – 2 = 0 (3x – 2)(x + 1) = 0

x = –1, 23

but if we divide all terms by x and manipulate it the equation at first glance does not appear as a quadratic 3x +1 =

2x

Yet this is an identical equation. Two other equations that can be solved as quadratic equations are 3x4 + x2 – 2 = 0

and 3x+ x = 2

In the first we substitute z = x2 to get our quadratic and in the second we substitute z = x .

ExampleSolvetheequation

3x+ x =2

Substitutez= x 3z2+z=2Rearrangesothattheequationequalszerofirst 3z2+z–2=0 Factorising (3z–2)(z+1) =0 Settingeachfactorequalto0gives z =–1,

23

x =z2Thiswouldappeartogiveanswers

x =1, 49

butsubstitutionintotheoriginalequationwefindtheonlyansweris x = 4

9Thisisbecausethesquarerootsignrepresentsthepositiverootonly(i.e.1)whichdoesnotsolvetheoriginalequation.Weshouldalwaysconfirmthesedisguisedquestionsbysubstitutingbackintotheoriginalequation.




407. (p+3)x2+4x+p=0hasequalroots.Find 408. Whatisthenatureoftherootsof thepossiblevaluesofp. (q+1)x2+(2q+1)x+q=0?

409. If(2k+3)x2–4kx+4=0hasequalroots findthevalue(s)ofk.

410. Findthevalue(s)ofkforwhich x2+(k+1)x=k+1hasequalroots.

411. Ifa,bandcarethreeconsecutivetermsofanarithmeticsequenceshowthattheequation (b–c)x2+(c–a)x+a–b=0hasequalroots.Usethediscriminantandshowallyourworking.




Practice External Assessment Task Algebraic Methods 2.6

Make sure you show ALL relevant working for each question.

You are advised to spend 60 minutes answering this assessment.

QUESTION ONE

(a) Expandandsimplify(2x+5)(x–3)(x–2)

(b) Writeasasinglelogstatement3logA+logB–0.5logC

(c) i) Solvetheequation3x2–4x+1=0

ii) Forwhatvalueofkdoestheequation3x2+kx+4=0haveequalroots?




Answers

Page 4

1. 4x2+20x2. 8x2–12x3. 2x3+3x2

4. 30a2–18a5. 6k3+4k2+2k6. 12x4+24x3–28x2

7. x2+8x+158. 3x2+2x–19. 6a2+5a+110. –30x2+25x+3011. 15x2–6x–1212. 12x2–9x+313. x2–2x–1214. 16x2–4x+1015. 21x2–14x16. x3–21x2+147x–34317. 12x3+24x2–3x–618. 3x3+x2–8x+419. a3+a2–14a–2420. 6x3+13x2–40x–7521. x3+x2–8x–1222. 6y3–19y2+y+623. x3–11x2+31x–2924. 64x3+48x2+12x+125. 3k3–16k2+32k–1726. –8y3+12y2–9y–1127. –12x2–16

Page 5

28. 8q3–12q2+6q–1

29. –11m2+5m

30. 6p3–7p2–6p+3

31. 12x2–28x+8

32. m3–7m–6

33. 24k3–58k2+23k+15

34. a)2x+40

b)πx2+40πx+400π

c) 40πx+400π d)x=5.92

Diameterofstageis 11.8m(1dp).

Page 735. 3x(7–2x)36. 5x(1–3x)37. x3yz(z+y+x)38. 3b(4a2+a–3b)39. 2πr(r+h)40. 7km(4k–7m)41. 4k2(1–3k)42. 12xy(3x–2y)43. (k–3)(4+y)44. (x+2)(x+7)45. (x–7)(x–1)46. (x–a)(4–2y) =2(x–a)(2–y)47. (x+2)(12x+a)48. (e+1)(3ac–6b) =3(e+1)(ac–2b)49. (4x–1)(2x+3z)

Page 850. (x+6)(x+1)51. (x+3)(x+4)52. x(x+9)53. (x–8)(x+3)54. (a+6)(a–4)55. (z+5)(z–5)56. (x+5)(x+1)57. (x+8)(x–3)58. (x–4)(x–8)59. (h+6)(h+6)=(h+6)2

60. (p+1)(p–1)61. (x+9)(x–9)62. (m–1)(m–6)63. (x–9)(x+5)64. (x+26)(x–2)

65. (m+7)(m–2)66. (k–9)(k–6)67. (r+7)(r–4)68. (a–9)(a+2)69. (p–6)2

70. (n+1)(n+12)71. (s–17)(s+4)72. (x+14)(x–8)73. (k+12)(k–9)74. (x+10)(x–10)75. (4k–7)(4k+7)76. (y–8x)(y+8x)

Page 9 cont...

77. (5–6a)(5+6a)78. (1–pq)(1+pq)

79. ( – )( )bbb

b1 1

+

80. (x–10)2

81. (x+8)2

82. (x+11)2

83. (x+13)2

84. (x–7)2

85. (x–15)2

Page 1086. 2(x–4)(x+1)87. 3(x–3)(x–1)88. 3(x+10)(x–1)89. (4b+5)(b+2)90. (x–3)(3x+2)91. (2n–5)(3n–4)92. 3(x–8)(x+5)93. 4(x+7)(x–2)94. 2(n+1)(n+12)

Page 1195. 3(p+7)(p–7)96. 5(q–7)(q–3)97. 0.5(n–8)(n+6)98. (3x+1)(x+1)99. (4x+1)(x–1)100. (5x+1)(x–2)101. (7x+3)(x+2)102. (3x–1)(3x+4)103. (2x+3)(3x–2)104. a)Length=60–2w Width=40–w

b)2w2–140w+2400=1408 w2–70w+496=0 c) w=8m(reject62) Length44m,Width32m

Page 14

105. 181

106. 2

107. 1125


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